Timeline for Non-Abelian fundamental group? --- a bizarre example
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16 events
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| Oct 28, 2019 at 10:27 | history | edited | YCor |
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| Sep 4, 2018 at 16:14 | comment | added | wonderich | Thanks, this is a harder question, but perhaps also doable: math.stackexchange.com/questions/2895658 | |
| Sep 4, 2018 at 16:11 | comment | added | YCor | I computed the inverse image of $G_1$ in the universal covering. This gives you the structure of the fundamental group. | |
| Sep 4, 2018 at 16:10 | comment | added | wonderich | I think the way to exclude $\pi_1(G)=\mathbb{Z}$ is due to that my eq (i) exclude that, even though eq (ii) allows it? So only $π_1(G)=ℤ×ℤ_N$ allowed ? | |
| Sep 4, 2018 at 16:06 | comment | added | YCor | $A\rtimes B$ means nothing if you don't specify the action of $B$ on $A$. Anyway, since $G$ is abelian, the only possibility is the direct product (which is one instance of semidirect product). By the way, if $N$ is odd, the only semidirect product $Z\rtimes (Z/NZ)$ is the direct product... | |
| Sep 4, 2018 at 15:44 | comment | added | wonderich | @YCor, $(not \mathbb{Z} \rtimes \mathbb{Z}_N?)$. | |
| Sep 4, 2018 at 15:44 | comment | added | wonderich | thanks, how we do we know it is $\pi_1(G)=\mathbb{Z}$ or $\pi_1(G)=\mathbb{Z} \times \mathbb{Z}_N$? It looks that both cases are possible. | |
| Sep 4, 2018 at 7:00 | review | Close votes | |||
| Sep 7, 2018 at 23:03 | |||||
| Sep 4, 2018 at 6:29 | comment | added | YCor | The fundamental group is isomorphic to the preimage $\tilde{G}_1$ of $G_1$ in the universal covering of $G_0$. This is indeed isomorphic to $Z\times (Z/nZ)$. | |
| Sep 4, 2018 at 3:51 | comment | added | Dan Ramras | As Ycor explained, general theory says the group in question is indeed Abelian. While there may be non-Abelian groups that can fit into the sequence (i), the group you're looking for is Abelian. | |
| Sep 3, 2018 at 20:45 | comment | added | wonderich | It can be, however, reading from the short exact sequence, from (i) - it does not seem to be an Abelian group but a non-Abelian group? from (ii), it can be an Abelian group? It is somehow puzzling. | |
| Sep 3, 2018 at 20:40 | comment | added | YCor | You mod out by a central subgroup, so the quotient is a Lie group, and in particular has an abelian fundamental group. | |
| Sep 3, 2018 at 20:29 | comment | added | wonderich | +1, thanks for the insightful ref - my example shall be a much easier example! | |
| Sep 3, 2018 at 19:17 | history | edited | wonderich | CC BY-SA 4.0 |
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| Sep 3, 2018 at 19:08 | history | edited | wonderich | CC BY-SA 4.0 |
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| Sep 3, 2018 at 19:00 | history | asked | wonderich | CC BY-SA 4.0 |